Question

尝试以这种方式使用sqlalchemy.schema.CheckConstraint：

themes2tags_table = Table('themes2tags', Base.metadata,
    Column('theme_id', String(32), ForeignKey('tags.id')),
    Column('tag_id', String(32), ForeignKey('tags.id')),
    PrimaryKeyConstraint(['theme_id', 'tag_id']),
    CheckConstraint("substr(theme_id,1,1)='3'"))

（意味着标签表上的多对多自引用关系，并且主题是具有预定义前缀的标签，显然），最终结果

    ...CheckConstraint("substr(theme_id,1,1)='3'")
  File "build\bdist.win32\egg\sqlalchemy\schema.py", line 303, in __new__
  File "build\bdist.win32\egg\sqlalchemy\schema.py", line 370, in _init
  File "build\bdist.win32\egg\sqlalchemy\schema.py", line 64, in _init_items
  File "build\bdist.win32\egg\sqlalchemy\events.py", line 234, in _set_parent_wi
th_dispatch
  File "build\bdist.win32\egg\sqlalchemy\schema.py", line 2133, in _set_parent
  File "build\bdist.win32\egg\sqlalchemy\schema.py", line 1909, in _set_parent
  File "build\bdist.win32\egg\sqlalchemy\schema.py", line 1883, in _set_parent
  File "build\bdist.win32\egg\sqlalchemy\sql\expression.py", line 2213, in add
AttributeError: 'list' object has no attribute 'key'

我在哪里错了？在这种情况下是否可能有这样的约束？

当前的后端是sqlite，但我考虑将来迁移到Postgres，所以寻找一个更不通用的收据。

更新：sqlalchemy 0.7.5

Answer 1

嗯，这是我的错，实际上是基于错误的错误信息。事实证明，实际错误发生在前一行，即正确的语法是：

themes2tags_table = Table('themes2tags', Base.metadata,
    Column('theme_id', String(32), ForeignKey('tags.id')),
    Column('tag_id', String(32), ForeignKey('tags.id')),
    PrimaryKeyConstraint(*['theme_id', 'tag_id']),
    CheckConstraint("substr(theme_id,1,1)='3'"))

在PrimaryKeyConstraint中注意列表定义之前的星号。

SQLAlchemy：CheckConstraint与字符串函数

1 个答案: